Is there a guarantee of stdout auto-flush before exit? How does it work?

问题

Here is the code (valid C and C++)

#include <stdio.h>

int main() {
    printf("asfd");
    // LINE 1
    return 0;
}

If in line 1 I put segfaulting expression the program would just crash without printing anything (as expected).

But why is the above code printing "asdf" and not exiting without buffer being flushed? What is under the hood and why does it work as expected?

回答1:

This is accomplished by these two sections in the C++ language specification:

[basic.start.main]

A return statement in main has the effect of leaving the main function and calling exit with the return value as the argument.

and

[lib.support.start.term]

The function exit has additional behavior in this International Standard:

• ...
• Next, all open C streams with unwritten buffered data are flushed.
• ...

回答2:

Generally, a return from main is not the end of your program, nor is entry to main the start.

Usually, the linker that creates the final executable for your program marks some location, perhaps named start, as the place where execution is to begin. When the operating system loads your program and starts executing it, it starts execution at this place. There is code there that sets up an environment: Creates a stack, sets stream states, et cetera. Then this code calls main.

When main returns, it returns to this special code. That code then performs various clean-up work that is required at the end of a C or C++ program, as described in this answer.

If a program is terminated abruptly, this final code might not be executed.

回答3:

When main() exits, all open streams are closed... to include stdout. Closing the open stream flushes stdout and what you've written to the buffer gets committed with or without the newline.

来源：https://stackoverflow.com/questions/15911517/is-there-a-guarantee-of-stdout-auto-flush-before-exit-how-does-it-work